What is the Value of v^2 in the Equation .0834 = .5 √(1 - v^2)?

[eNathan]

.0834 = .5 \sqrt { 1 - v^2 }

What is v^2?

 

[dextercioby]

Square both terms and isolate v^2...

Daniel.

 

[eNathan]

can you explain?

 

[dextercioby]

(0.0834)^2 =(0.5)^2 \left(1-v^2 \right)

Divide by 0.25.

Daniel.

 

[mathwonk]

it seems to be about 35/36 just off the top of my head.

 

[HallsofIvy]

can you explain?

Explain what? Do you know how to square a number? Do you know what \(\sqrt{1- x^2}\)^2 is?

Although I would be inclined to multiply both sides of the equation by 2 first.

 

[JoshHolloway]

I would have divided both sides by 1/2 first. then squared both sides.

 

[eNathan]

Ok, let me ask it in a simpler form.

If L_t = \sqrt {1 - \frac {v^2} {c^2} } then what does v^2 equal? BTW is it posible for the explination to be in the form of an anctual equation?

Thx

 

[dextercioby]

There's no explanation that could be given by an equation.Explaining means putting words,describing a method...


Daniel.

 

[dx]

c^2 ( 1 - L^2 )

 

[eNathan]

There's no explanation that could be given by an equation.Explaining means putting words,describing a method...


Daniel.

you can explain things using an equation by studying it, hence understanding it.

In any case, I have one more questionair. It seems I have forgot to include the LT factor in the equation.

If T = t \sqrt {1- \frac {v^2} {c^2} } then what is v^2

 

[eNathan]

Ok, I think it is
v^2 = c^2 \frac {(1-T^2)} {t}

 

[jdavel]

Ok, I think it is
v^2 = c^2 \frac {(1-T^2)} {t}

Nope. Try again.

 

[mathwonk]

another casualty of the epidemic decline of algebra skills since the introduction of the hand held calculator, and MTV, and the republican controlled congress, and daylight savings time, and fluoridation of the water, and the new deal, and...oops time for my medication.

 

[eNathan]

J338!5 I though physicsforums.com was to help people, not talk about the decline in inteligence.

--Nuf Said

 

[TenaliRaman]

eNathan,
the simplified equation you were after was
v^2 = c^2 (1 - (T^2/t^2))

-- AI

 

[whozum]

another casualty of the epidemic decline of algebra skills since the introduction of the hand held calculator, and MTV, and the republican controlled congress, and daylight savings time, and fluoridation of the water, and the new deal, and...oops time for my medication.

You forgot to mention the introduction of baby pickles!

 

[whozum]

Ok, let me ask it in a simpler form.

If L_t = \sqrt {1 - \frac {v^2} {c^2} } then what does v^2 equal? BTW is it posible for the explination to be in the form of an anctual equation?

Thx

Since the answer has been provided I'll show you how.

T = t\sqrt{1-\frac{v^2}{c^2}}

Divide both sides by t

\frac{T}{t} = \sqrt{1-\frac{v^2}{c^2}}

Square both sides

\left(\frac{T}{t}\right)^2 = 1-\frac{v^2}{c^2}

Subtract one from both sides, then since the right side is negative, multiply both sides by negative one.

1-\left(\frac{T}{t}\right)^2 = \frac{v^2}{c^2}

Multiply both sides by c^2

c^2\left(1 - \frac{T^2}{t^2}\right) = v^2

 

[eNathan]

Since the answer has been provided I'll show you how.

T = t\sqrt{1-\frac{v^2}{c^2}}

Divide both sides by t

\frac{T}{t} = \sqrt{1-\frac{v^2}{c^2}}

Square both sides

\left(\frac{T}{t}\right)^2 = 1-\frac{v^2}{c^2}

Subtract one from both sides, then since the right side is negative, multiply both sides by negative one.

1-\left(\frac{T}{t}\right)^2 = \frac{v^2}{c^2}

Multiply both sides by c^2

c^2\left(1 - \frac{T^2}{t^2}\right) = v^2

Derived! Makes perfect sense, thanks a lot for the explinationair

 

[whozum]

You should definitely brush up on your algebra. It's essential to your science career.

 

[eNathan]

Actaully this is what I originally derived

V^2 = -c^2 ( \frac{T'} {t} )^2 - 1

BTW I am taking algebra II right now.

 

[dextercioby]

Maybe he'll go into poetry... Why should he bother with algebra...?

Take square roots and cross multiply the rhymes...?

Daniel.

 

[mathwonk]

dear nuff sedd: that would be "intelligence".


har har har.

physics forum is also intended to provide amusement at the expense of random posters.

sincerely,

the gerbil.

i just googled "pikachu", and found some "kill pikachu" video game site, so i presume he is kind of a gameboy version of al capp's "schmoos", i.e. little harmless creatures so tame thye let themselves be chopped up into ham and eggs for the benefit of hungry hillbillies. helloo? anybody heard of al capp?

daisy mae? it was sort of a gilligan's island on the comic page only cleverer, and set in appalachia. sadie hawkins day?

next you are going to claim you don't know who pogo is either. and you expect to become physicists?

by the way have you noticed that asperger's syndrome is an occupational hazard of mathematicians?

our patron saint is detective monk, of the tv show.

well what's so strange about touching all the parking meters? I've always done that! In fact I used to walk on the tops of them, but that got harder.

 

[whozum]

mw: Your starting to scare me man.
dex: poetry could use some new flavor :)

enathan: sorry, i was under the impression you were a college student

 

[eNathan]

Ok, can someone please tell me what's wrong with this

We have the original equation
T = t \sqrt { 1- \frac {v^2} {c^2} }

Therefore

\frac {T} {t} = \sqrt { 1- \frac {v^2} {c^2} }

Because the "square root" is on the right, we can square the left

(\frac {T} {t})^2 = 1- \frac {v^2} {c^2}

And we all know that 1-x is the same as -x+1 -- I hope I am right on this one

(\frac {T} {t})^2 = \frac {-v^2} {c^2} +1

Because we add +1 on the right side of the equation, we add -1 on the left

(\frac {T} {t})^2 - 1 = \frac {-v^2} {c^2}

Because we divide by c^2 on the right of the equation, we can multiply it on the left.

c^2 (\frac {T} {t})^2 - 1 = -v^2

Because we have -1(v^2) on the right, we can divide both sides of the equation by -1 which yeilds a -c^2 on the left

-c^2 (\frac {T} {t})^2 - 1 = v^2

And of course, to get v by itself we just do

\sqrt { -c^2 (\frac {T} {t})^2 - 1 } = v

Now I know I am not too 600[) at algebra, but I just do not see what is wrong here.

Any help?

 

[snoble]

(\frac {T} {t})^2 - 1 = \frac {-v^2} {c^2}

Because we divide by c^2 on the right of the equation, we can multiply it on the left.

c^2 (\frac {T} {t})^2 - 1 = -v^2

It's the law of distributivity. If you multiply the left hand side by c^2 you get

c^2\cdot ((\frac {T} {t})^2 - 1)= c^2(\frac {T} {t})^2- c^2

Think of math like laTex. You don't have to put everything in brackets but if you do it shouldn't change things. When you multiply the left hand side by something you should get the same result whether it is in brackets or it is not. I'm just saying that if you can write tex you can use algebra. Good tex is harder than good algebra.

Steven

 

[master_coda]

It's the law of distributivity. If you multiply the left hand side by c^2 you get

c^2\cdot ((\frac {T} {t})^2 - 1)= c^2(\frac {T} {t})^2- c^2

Think of math like laTex. You don't have to put everything in brackets but if you do it shouldn't change things. When you multiply the left hand side by something you should get the same result whether it is in brackets or it is not. I'm just saying that if you can write tex you can use algebra. Good tex is harder than good algebra.

Steven

You also made the same mistake when you divided by -1. You only divided the first term of the left hand side, not both of them.

 

[eNathan]

Is this correct then?

\sqrt { -c^2 \cdot ((\frac {T} {t})^2 - 1) } = v

 

[Moo Of Doom]

Yes, exactly.

 

[eNathan]

Well I guess I wasnt far off, I just overlooked the order of operations.

Thanks everyone

 

[floped perfect]

v^2= [c^2(t^2-T^2)}/t^2

 

[HallsofIvy]

Alright, I'll bite: what does "floped" mean?